Let's consider some cases:
1) the first part of the answer is a min number of goals needed to have a chance to advance next round. That means that the second team will score 0 goals in second game, and how many goals needed to be scored to complete the condition??
2) the second part - max number of goals to be scored in such case, that the opposite team still will have a change to win. That means we must suppose second team to have scored 30 goals in second round.

in each case we must calculate number of our scores to make the summarized scores equal for 2 rounds. But then we check the condition if the numbers of goals scored on the away fields of each team are equal - then the answer is OK, every team has a chance to advance. In first case we must take ++value of number if quantity of scored goals on away team is less (that means no chance), in second - value-- of number if quantity of scored goals on away team is greater(cause that leaves no chance for opposite team to win)

is that enough to solve??

why wa2???
program ural;
var n,m,i,k,j,r,p:byte;
    a:array[1..100]of byte;
    b:array[1..100,1..100]of byte;
    d:array[1..100]of integer;
    z,t:integer;
label 1;
begin
  readln(n,m);
  for i:=1 to m do
    begin
      read(k);
      for j:=1 to k do
        read(a[j]);
      for j:=1 to k do
        for r:=1 to k do
          if j<>r then
            begin
              b[a[j],a[r]]:=1;
              b[a[r],a[j]]:=1;
            end;
    end;
  readln(m);
  for i:=1 to m do
    begin
      readln(d[i],a[i],p);
      if p=1 then
        d[i]:=101
      else
        d[i]:=d[i] div 4;
    end;
  for i:=1 to n do
    for j:=1 to n-1 do
      if i<>j then
        for r:=j+1 to n do
          if r<>i then
            if(b[j,i]>0)and(b[i,r]>0)then
              begin
                t:=b[j,i]+b[i,r];
                if(b[j,r]=0)or(t<b[j,r])then
                  begin
                    b[j,r]:=t;
                    b[r,j]:=t;
                  end;
              end;
  z:=255;
  for i:=1 to n do
    begin
      t:=0;
      for j:=1 to m do
        if a[j]<>i then
          if(b[a[j],i]>0)and(b[a[j],i]<=d[j])then
            begin
              if d[i]<>101 then
                inc(t,b[a[j],i]);
            end
          else
            goto 1;
      if t<z then
        begin
          k:=i;
          z:=t;
        end;
1:  end;
  if z=255 then
    writeln(0)
  else
    writeln(k,' ',z*4);
end.

The Machinegunners played away game, scored 0 goals, and conceded 0 goals.
The Machinegunners played away game, scored 15 goals, and conceded 15 goals.

ans
0 30
0 30

Edited by author 02.11.2010 20:39

Please, write somebody full answer for N = 994009.

can anybody give me another sample?what is the mean of "Output limit exceeded"?

This test: 900660121

My solution printed 30011 30011 30011 510187 570209 690253 870319 930341 1110407 1230451 1290473 1410517 1470539 1590583 1770649 1830671 2010737 2130781 2190803 2370869 2430891 2490913 2670979 2911067 3031111 3091133 3211177 3271199 3391243 3631331 3751375 3811397 3841408 3931441 4111507 4171529 4471639 4531661 4711727 4891793 5011837 5071859 5191903 5371969 5431991 5732101 5792123 5912167 5972189 6332321 6692453 6812497 6872519 6992563 7172629 7232651 7532761 7712827 7892893 8072959 8132981 8313047 8433091 8493113 8793223 9213377 9333421 9393443 9513487 9933641 10113707 10413817 10473839 10593883 10773949 11014037 11194103 11374169 11494213 11674279 11914367 12034411 12274499 12574609 12634631 12934741 12994763 13174829 13294873 13474939 13715027 13835071 13895093 14015137 14375269 14615357 14735401 14975489 15095533 15275599 15695753 16235951 16416017 16716127 16896193 17076259

But I think that first two 30011 is better to substitute for 60022. Or where am I "stuping"? =)

В обоих случаях из примера бутерброд за время падения не совершает ни 1 полного оборота, соответственно он падает в обоих случаях маслом вниз.

Скорость вращения - 1/4 оборота в секунду.

Время падения 1 раз = 0,9.
Количество оборотов 1 раз = 0,2258.

Время падения 2 раз = 1,01.
Количество оборотов 2 раз = 0,2524.

В обоих случаях бутерброд НЕ совершает ни 1 полного оборота и должен упасть как и был, маслом вниз.

Почему же ответы для 1 примера - "маслом вниз", а для 2 примера - "хлебом вниз"?

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>

using namespace std;

void recursive_print(){
    double A; char ch;
    if ((ch = getchar()) != EOF){
        cin >> A;
        recursive_print();
        cout << endl << setiosflags(ios::showpoint | ios::fixed) << setprecision(4) << sqrt(A);
    }
}

int main(){
    recursive_print();
}

test:
0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0
3
White
a1-a2
h8-h7
a2-a4

ans:
Correct
Correct
Correct

0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0

#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main()
{
  int n,k,mh=0,otg=999;
  string s;
  vector<int>v;
  cin>>k;
  for(int i=1;i<=k;i++)
  {
    cin>>n;
    cin>>s;
    if(s=="hungry")
    {
      if(n>mh) mh=n;
    }
    else v.push_back(n);
  }
  if(v.size())
  {
    for(int i=0;i<v.size();i++)
    {
      if(v[i]>mh&&v[i]<otg) otg=v[i];
    }
    if(otg==999) cout<<"Inconsistent"<<endl;
    else cout<<otg<<endl;
  }
  else cout<<"10"<<endl;
  return 0;
}

Edited by author 02.11.2010 12:52

Edited by author 02.11.2010 12:52

Which method can be used in this problem? Really dynamic programming? Give me this test if you can please... :)

#include <iostream>

using namespace std;

int main()
{
    long int a, i=0, n, k1=0, k2=0;

    cin>>n;

    while( i < n )
    {
        cin>>a;
        if( k1 < k2 )
            k1 += a;
        else
            k2 += a;
        i++;
    }

    if( k1 >= k2 )
        cout<<k1-k2<<endl;
    else
        cout<<k2-k1<<endl;

    return 0;
}

 I need help because in the test 6 I have a crash and I do not
 understand why?
 And I'm sure that the algorithm to this problem is correct,
 Thanks..

What is the answer for this tests?

The Machinegunners played away game, scored 1 goals, and conceded 2 goals.

The Machinegunners played home game, scored 0 goals, and conceded 1 goals.

The Machinegunners played away game, scored 1 goals, and conceded 1 goals.

Can they look though each other?

What is the answer for:
The Machinegunners played away game, scored 30 goals, and conceded 0 goals.
?

If they loose 0 : 30, they advance to the next round anyway, am I right?

I have Wa #8 many times, I don't know what's wrong!!! Can anybody check my code or give me a hint?

can u give me some tests.. plz

You should add this to get AC:

if (ans < 0) ans = 0;
if (ans > 30) ans = 30;